Question

If ax+by/x+y=bx+az/x+z=ay+bz/y+z and x+y+z≠0 then show that a+b/2

Answer 1

Answer:

$\frac{ax+by}{x+y}=\frac{bx+az}{x+z}=\frac{ay+bz}{y+z}=\frac{(a+b)}{2}$

Step-by-step explanation:

If ax+by/x+y=bx+az/x+z=ay+bz/y+z and x+y+z≠0 then show that a+b/2

Concept used:

$\boxed{\frac{a}{x}=\frac{b}{y}=\frac{c}{z}\:\implies\:\frac{a}{x}=\frac{b}{y}=\frac{c}{z}=\frac{a+b+c}{x+y+z}}$

$\frac{ax+by}{x+y}=\frac{bx+az}{x+z}=\frac{ay+bz}{y+z}$

$\implies\:\frac{ax+by}{x+y}=\frac{bx+az}{x+z}=\frac{ay+bz}{y+z}=\frac{ax+by+bx+az+ay+bz}{x+y+x+z+y+z}$

$\implies\:\frac{ax+by}{x+y}=\frac{bx+az}{x+z}=\frac{ay+bz}{y+z}=\frac{a(x+y+z)+b(x+y+z)}{2(x+y+z)}$

$\implies\:\frac{ax+by}{x+y}=\frac{bx+az}{x+z}=\frac{ay+bz}{y+z}=\frac{(a+b)(x+y+z)}{2(x+y+z)}$

$\implies\:\boxed{\frac{ax+by}{x+y}=\frac{bx+az}{x+z}=\frac{ay+bz}{y+z}=\frac{(a+b)}{2}}$